[**Revised and expanded** to give the answer for all $k>1$ and incorporate
further terms of an asymptotic expansion as $n \rightarrow \infty$]

Fix $k>1$, and write $a_1=f(1,k)=1$ and
$$
a_n = f(n,k) =
\frac1{1-q^{-n}} \sum_{r=1}^{n-1} {n \choose r} (1/k)^{n-r} (1/q)^r a_r
\phantom{for}(n>1),
$$
where $q := k/(k-1)$, so $(1/k) + (1/q) = 1$. Set
$$
a_\infty := \frac1{k \log q}.
$$
For example, if $k=2$ then $a_\infty = 1 / \log 4 = 0.72134752\ldots$,
which $a_n$ seems to approach for large $n$, and likewise for
$k=6$ (the dice-throwing case) with $a_\infty = 1/(6 \log 1.2) =
0.9141358\ldots$. Indeed as $n \rightarrow \infty$ we have
"$a_n \rightarrow a_\infty$ on average", in the sense that (for instance)
$\sum_{n=1}^N (a_n/n) \sim a_\infty \phantom. \sum_{n=1}^N (1/n)$
as $N \rightarrow \infty$. But, as suggested by earlier
posted answers to Tim Chow's question, $a_n$ does not converge,
though it stays quite close to $a_\infty$: we have
$$
a_n = a_\infty + \epsilon^{\phantom.}_0(\log_q n) + O(1/n)
$$
as $n \rightarrow \infty$, where $\epsilon^{\phantom.}_0$ is a smooth
function of period $1$ whose average over ${\bf R} / {\bf Z}$ vanishes
but is not identically zero; for large $k$ (already $k=2$ is large
enough), $\epsilon^{\phantom.}_0$ is a nearly perfect sine wave
with a tiny amplitude $\exp(-\pi^2 k + O(\log k))$, namely
$$
\frac2{k\log q}\left|\phantom.\Gamma\bigl(1 + \frac{2\pi i}{\log q}\bigr)\right|
\phantom.=\phantom.
\frac2{k \log q}
\left[\frac{(2\pi^2/ \log q)}{\sinh(2\pi^2/ \log q)}\right]^{1/2}.
$$
For example, for $k=2$ the amplitude is $7.130117\ldots \cdot 10^{-6}$,
in accordance with numerical observation (see previously posted answers
and the plot below). For $k=6$ the amplitude is only
$8.3206735\ldots \cdot 10^{-23}$ so one must compute well beyond
the usual "double precision" to see the oscillations.

More precisely, there is an asymptotic expansion
$$
a_n \sim a_\infty + \epsilon^{\phantom.}_0(\log_q n)
+ n^{-1} \epsilon^{\phantom.}_1(\log_q n)
+ n^{-2} \epsilon^{\phantom.}_2(\log_q n)
+ n^{-3} \epsilon^{\phantom.}_3(\log_q n)
+ \cdots,
$$
where each $\epsilon^{\phantom.}_j$ is smooth function of period $1$
whose average over ${\bf R} / {\bf Z}$ vanishes, and
— while the series need not converge —
truncating it before the term $n^{-j} \epsilon^{\phantom.}_j(\log_q n)$
yields an approximation good to within $O(n^{-j})$. The first few
$\epsilon^{\phantom.}_j$ still have exponentially small amplitudes,
but larger that of $\epsilon^{\phantom.}_0$ by a factor
$\sim C_j k^{2j}$ for some $C_j > 0$; for instance, the amplitude of
$\epsilon^{\phantom.}_1$ exceeds that of $\epsilon^{\phantom.}_0$
by about $2(\pi / \log q)^2 \sim 2 \pi^2 k^2$. So $a_n$ must be
computed up to a somewhat large multiple of $k^2$ before it becomes
experimentally plausible that the residual oscillation $a_n - a_\infty$
won't tend to zero in the limit as $n \rightarrow \infty$.

Here's a plot that shows $a_n$ for $k=2$ (so also $q=2$) and
$2^6 \leq n \leq 2^{13}$, and compares with the periodic approximation
$a_\infty + \epsilon^{\phantom.}_0(\log_q n)$ and the refined approximation
$a_\infty + \sum_{j=0}^2 n^{-j} \epsilon^{\phantom.}_j(\log_q n)$.
(See http://math.harvard.edu/~elkies/mo11255+.pdf for
the original PDF plot, which can be "zoomed in" to view details.) The
horizontal coordinate is $\log_2 n$; the vertical coordinate is centered at
$a_\infty = 1/(2 \log 2)$, showing also the lines $a_\infty \pm 2|a_1|$;
black cross-hairs, eventually merging visually into a continuous curve,
show the numerical values of $a_n$; and the red and green contours
show the smooth approximations.

To obtain this asymptotic expansion, we start by generalizing
R.Barton's formula from $k=2$ to arbitrary $k>1$:
$$
a_n = \frac1k \sum_{r=0}^\infty \phantom. n q^{-r} (1-q^{-r})^{n-1}.
$$
[The proof is the same, but note the exponent $n$ has been corrected
to $n-1$ since we want $n-1$ players eliminated at the $r$-th step,
not all $n$; this does not affect the limiting behavior
$a_\infty+\epsilon^{\phantom.}_0(\log_q n)$, but is needed to get
$\epsilon^{\phantom.}_m$ right for $m>1$.] We would like to approximate
the sum by an integral, which can be evaluated by the change of variable
$q^{-r} = z$:
$$
\frac1k \int_{r=0}^\infty \phantom. n q^{-r} (1-q^{-r})^{n-1}
= \frac1{k \log q} \int_0^1 \phantom. n (1-z)^{n-1} dz
= \left[-a_\infty(1-z)^n\right]_{z=0}^1 = a_\infty.
$$
But it takes some effort to get at the error in this approximation.

We start by comparing $(1-q^{-r})^{n-1}$ with $\exp(-nq^{-r})$:
$$
\begin{eqnarray}
(1-q^{-r})^{n-1}
&=&
\exp(-nq^{-r}) \cdot \exp \phantom. [nq^{-r} + (n-1) \log(1-q^{-r})]
\cr
&=&
\exp(-nq^{-r}) \cdot
\exp \left[q^{-r} - (n-1) \left(
\frac{q^{-2r}}2 + \frac{q^{-3r}}3 + \frac{q^{-4r}}4 + \cdots
\right)
\right].
\end{eqnarray}
$$
The next two steps require justification (as R.Barton noted for
the corresponding steps at the end of his analysis), but the
justification should be straightforward. Expand the second factor
in powers of $u := nq^{-r}$, and collect like powers of $n$, obtaining
$$
\exp(-nq^{-r}) \cdot \left(
1 - \frac{u^2-2u}{2n} + \frac{3u^4-20u^3+24u^2}{24n^2}
- \frac{u^6-14u^5+52u^4-48u^3}{48n^3} + - \cdots \right).
$$
Each term $n^{-j} \epsilon_j(\log_q(n))$ ($j=0,1,2,3,\ldots$)
will arise from the $n^{-j}$ term in this expansion.

We start with the main term, for $j=0$,
which is the only one that does not decay with $n$. Define
$$
\varphi_0(x) := q^x \exp(-q^x),
$$
which as Reid observed decays rapidly both as $x \rightarrow \infty$
and as $x \rightarrow -\infty$. Our zeroth-order approximation to $a_n$ is
$$
\frac1k \sum_{r=0}^\infty \phantom. \varphi_0(\log_q(n)-r),
$$
which as $n \rightarrow \infty$ rapidly nears
$$
\frac1k \sum_{r=-\infty}^\infty \varphi_0(\log_q(n)-r).
$$
For $k=q=2$, this is equivalent with Reid's formula for $R(n)$,
even though he wrote it in terms of the fractional part of $\log_2(n)$,
because the sum is clearly invariant under translation of $\log_q(n)$
by integers.

We next apply Poisson summation. Since
$\sum_{r \in {\bf Z}} \phantom. \varphi_0(t+r)$
is a smooth ${\bf Z}$-periodic function of $t$, it has a Fourier expansion
$$
\sum_{m\in{\bf Z}} \phantom. c_m e^{2\pi i m t}
$$
where
$$
c_m = \int_0^1 \left[ \sum_{r \in {\bf Z}} \phantom. \varphi_0(t+r) \right]
\phantom. e^{-2\pi i m t} dt
= \int_{-\infty}^\infty \varphi_0(t) \phantom. e^{-2\pi i m t} dt
= \hat\varphi_0(-m).
$$
Changing the variable of integration from $t$ to $q^t$ lets us recognize the
Fourier transform $\hat\varphi$ as $1/\log(q)$ times a Gamma integral:
$$
\hat\varphi_0(y)
= \frac1{\log q} \Gamma\Bigl(1 + \frac{2 \pi i y} {\log q}\Bigr).
$$
This gives us the coefficients $a_m$ in closed form. The constant coefficient
$a_0 = 1 / (\log q)$ can again be interpreted as the approximation of the
Riemann sum $\sum_{r \in {\bf Z}} \phantom. \varphi_0(t+r)$ by an integral;
the oscillating terms $a_m e^{2\pi i m t}$ for $m \neq 0$
are the corrections to this approximation, and are small due to the
exponential decay of the Gamma function on vertical strips — indeed
we can compute the magnitude $|a_m|$ in elementary closed form using the formula
$|\Gamma(1+i\tau)| = (\pi\tau / \sinh(\pi\tau))^{1/2}$. So we have
$$
\frac1k \sum_{r \in \bf Z} \phantom. \varphi_0(\log_q(n)-r)
= \frac1k \sum_{m \in \bf Z} \phantom. \hat\varphi_0(-m) e^{2\pi i \log_q(n)}
= a_\infty + \epsilon_0(\log_q(n))
$$
where $a_\infty = a_0 / k = 1 / (k \log q)$ as above, and
$\epsilon^{\phantom.}_0$, defined by
$$
\epsilon^{\phantom.}_0(t) =
\left[ \sum_{r\in\bf Z} \phantom. \varphi_0(t+r) \right] - a_\infty,
$$
has the Fourier series
$$
\epsilon^{\phantom.}_0(t)
= \frac1k \sum_{m \neq 0} \hat\varphi_0(-m) e^{2\pi i m t}.
$$
Taking $m = \pm 1$ recovers the amplitude $2|a_1|/k$ exhibited above;
the $m = \pm 2$ and further terms yield faster but tinier oscillations,
e.g. for $k=2$ the $m=\pm 2$ terms oscillate twice as fast but with
amplitude only $6.6033857\ldots \cdot 10^{-12}$.

The functions $\epsilon^{\phantom.}_j$ appearing in the further terms
$n^{-j} \epsilon^{\phantom.}_j(\log_q(n))$
of the asymptotic expansion of $a_n$ are defined similarly by
$$
\epsilon^{\phantom.}_j(t) =
\frac1k \sum_{r\in\bf Z} \phantom. \varphi_j(t+r),
$$
where
$$
\varphi_j(x) = P_j(q^x) \varphi_0(x) = P_j(q^x) q^x \exp(-q^x)
$$
and $P_j$ is the coefficient of $n^{-j}$ in our power series
$$
(1-q^r)^{n-1} = \exp(-nq^{-r}) \phantom.
\sum_{j=0}^\infty \frac{P_j(nq^{-r})}{n^j}.
$$
Thus
$
P_0(u)=1, \phantom+
P_1(u) = -(u^2-2u)/2, \phantom+
P_2(u) = (3u^4-20u^3+24u^2)/24
$, etc.
Again we apply Poisson to expand $\epsilon^{\phantom.}_j(\log_q(n))$
in a Fourier series:
$$
\epsilon^{\phantom.}_j(t)
= \frac1k \sum_{m \in \bf Z} \hat\varphi_j(-m) e^{2\pi i m t},
$$
and evaluate the Fourier transform $\hat\varphi_j$ by integrating
with respect to $q^t$. This yields a linear combination of Gamma
integrals evaluated at $1 + (2\pi i y / \log q) + j'$ for
integers $j' \in [j,2j]$, giving $\hat\varphi_j$ as a
degree-$2j$ polynomial multiple of $\hat\varphi_0$. The first case is
$$
\begin{eqnarray*}
\hat\varphi_1(y)
&=& \frac1{\log q} \left[
\Gamma\Bigl(2 + \frac{2 \pi i y} {\log q}\Bigr)
- \frac12 \Gamma\Bigl(3 + \frac{2 \pi i y} {\log q}\Bigr)
\right]
\\
&=& \frac1{\log q} \frac{\pi y}{\log q} \left(\frac{2 \pi y}{\log q} - i\right)
\phantom.
\Gamma\Bigl(1 + \frac{2 \pi i y} {\log q}\Bigr)
\\
&=& \frac{\pi y}{\log q} \left(\frac{2 \pi y}{\log q} - i\right)
\phantom. \hat\varphi_0(y).
\end{eqnarray*}
$$
Note that $\varphi_1(0) = 0$, so the constant coefficient of the
Fourier series for $\epsilon^{\phantom.}_1(t)$ vanishes;
this is indeed true of $\epsilon^{\phantom.}_j(t)$ for each $j>0$,
because $\hat\varphi_j(0) = \int_{-\infty}^\infty \phi_j(x) \phantom. dx$
is the $n^{-j}$ coefficient of a power series in $n^{-1}$ that we've
already identified with the constant $a_\infty$. Hence (as can also
be seen in the plot above) none of the decaying corrections
$n^{-j} \epsilon^{\phantom.}_j(\log_q n)$ biases the average of $a_n$
away from $a_\infty$, even when $n$ is small enough that
those corrections are a substantial fraction of the
residual oscillation $\epsilon_0(\log_q n)$. This leaves
$\hat\varphi_j(\mp1) e^{\pm 2 \pi i t} / k$ as the leading terms
in the expansion of each $\epsilon^{\phantom.}_j(t)$, so we see as promised
that $\epsilon^{\phantom.}_j$ is still exponentially small but with
an extra factor whose leading term is a multiple of $(2\pi / \log q)^{2j}$.